(\large\bf Function derivative)
Consider the function y=f(x), given on the interval (a,b). Let x- any fixed point interval (a,b), A Δx- an arbitrary number, such that the value x+Δx also belongs to the interval (a,b). This number Δx is called argument increment.
Definition. Function increment y=f(x) at the point x, corresponding to the increment of the argument Δx, let's call the number
Δy = f(x+Δx) - f(x).
We believe that Δx ≠ 0. Consider at a given fixed point x the ratio of the increment of the function at that point to the corresponding increment of the argument Δx
This relation will be called the difference relation. Since the value x we consider fixed, the difference relation is a function of the argument Δx. This function is defined for all argument values Δx, belonging to some sufficiently small neighborhood of the point ∆x=0, except for the point ∆x=0. Thus, we have the right to consider the question of the existence of a limit of the specified function for ∆x → 0.
Definition. Derivative function y=f(x) at a given fixed point x is called the limit ∆x → 0 differential relation, that is
Provided that this limit exists.
Designation. y (x) or f′(x).
The geometric meaning of the derivative: Derivative of function f(x) at this point x equal to the tangent of the angle between the axis Ox and a tangent to the graph of this function at the corresponding point:
f′(x 0) = \tgα.
The mechanical meaning of the derivative: The derivative of the path with respect to time is equal to the speed of the rectilinear movement of the point:
Line tangent equation y=f(x) at the point M0 (x0,y0) takes the form
y-y 0 = f (x 0) (x-x 0).
The normal to the curve at some point is the perpendicular to the tangent at the same point. If f′(x 0)≠ 0, then the equation of the normal to the line y=f(x) at the point M0 (x0,y0) is written like this:
The concept of differentiability of a function
Let the function y=f(x) defined on some interval (a,b), x- some fixed value of the argument from this interval, Δx- any increment of the argument such that the value of the argument x+Δx ∈ (a, b).
Definition. Function y=f(x) is called differentiable at a given point x if increment Δy this function at the point x, corresponding to the increment of the argument Δx, can be represented as
Δy = A Δx +αΔx,
Where A is some number independent of Δx, A α - argument function Δx, which is infinitely small at ∆x → 0.
Since the product of two infinitesimal functions αΔx is an infinitesimal higher order than Δx(property 3 of infinitesimal functions), we can write:
∆y = A ∆x +o(∆x).
Theorem. In order for the function y=f(x) was differentiable at a given point x, it is necessary and sufficient that it has a finite derivative at this point. Wherein A=f′(x), that is
Δy = f′(x) Δx +o(Δx).
The operation of finding the derivative is usually called differentiation.
Theorem. If the function y=f(x) x, then it is continuous at that point.
Comment. From the continuity of the function y=f(x) at this point x, generally speaking, it does not follow that the function is differentiable f(x) at this point. For example, the function y=|x|- continuous at a point x=0, but has no derivative.
The concept of a function differential
Definition. function differential y=f(x) is called the product of the derivative of this function and the increment of the independent variable x:
dy = y′ ∆x, df(x) = f′(x) ∆x.
For function y=x we get dy=dx=x'Δx = 1 Δx= Δx, that is dx=Δx- the differential of an independent variable is equal to the increment of this variable.
Thus, we can write
dy = y′dx, df(x) = f′(x)dx
Differential dy and increment Δy functions y=f(x) at this point x, both corresponding to the same increment of the argument Δx are, in general, not equal to each other.
The geometric meaning of the differential: The differential of a function is equal to the increment of the ordinate of the tangent to the graph of the given function when the argument is incremented Δx.
Differentiation rules
Theorem. If each of the functions u(x) And v(x) differentiable at a given point x, then the sum, difference, product and quotient of these functions (quotient provided that v(x)≠ 0) are also differentiable at this point, and the following formulas hold:
Consider a complex function y=f(φ(x))≡ F(x), Where y=f(u), u=φ(x). In this case u called intermediate argument, x - independent variable.
Theorem. If y=f(u) And u=φ(x) are differentiable functions of their arguments, then the derivative of the complex function y=f(φ(x)) exists and is equal to the product of this function with respect to the intermediate argument and the derivative of the intermediate argument with respect to the independent variable, i.e.
Comment. For a complex function that is a superposition of three functions y=F(f(φ(x))), the differentiation rule has the form
y′ x = y′ u u′ v v′ x,
where functions v=φ(x), u=f(v) And y=F(u) are differentiable functions of their arguments.
Theorem. Let the function y=f(x) is increasing (or decreasing) and continuous in some neighborhood of the point x0. Let, in addition, this function be differentiable at the indicated point x0 and its derivative at this point f′(x 0) ≠ 0. Then in some neighborhood of the corresponding point y0=f(x0) the inverse for y=f(x) function x=f -1 (y), and the indicated inverse function is differentiable at the corresponding point y0=f(x0) and for its derivative at this point y the formula is valid
Derivative table
Invariance of the form of the first differential
Consider the differential of a complex function. If y=f(x), x=φ(t) are differentiable functions of their arguments, then the derivative of the function y=f(φ(t)) is expressed by the formula
y′ t = y′ x x′ t.
A-priory dy=y't dt, then we get
dy = y′ t dt = y′ x x′ t dt = y′ x (x′ t dt) = y′ x dx,
dy = y′ x dx.
So, we have proven
Property of invariance of the form of the first differential of a function: as in the case when the argument x is an independent variable, and in the case when the argument x is itself a differentiable function of the new variable, the differential dy functions y=f(x) is equal to the derivative of this function, multiplied by the differential of the argument dx.
Application of the differential in approximate calculations
We have shown that the differential dy functions y=f(x), generally speaking, is not equal to the increment Δy this function. Nevertheless, up to an infinitely small function of a higher order of smallness than Δx, the approximate equality
∆y ≈ dy.
The ratio is called the relative error of the equality of this equality. Because ∆y-dy=o(∆x), then the relative error of this equality becomes arbitrarily small as |Δх|.
Given that Δy=f(x+δx)-f(x), dy=f′(x)Δx, we get f(x+δx)-f(x) ≈ f′(x)Δx or
f(x+δx) ≈ f(x) + f′(x)Δx.
This approximate equality allows with an error o(Δx) replace function f(x) in a small neighborhood of a point x(i.e. for small values Δx) a linear function of the argument Δx standing on the right side.
Derivatives of higher orders
Definition. The second derivative (or second order derivative) of the function y=f(x) is called the derivative of its first derivative.
Notation for the second derivative of a function y=f(x):
Mechanical meaning of the second derivative. If the function y=f(x) describes the law of motion of a material point in a straight line, then the second derivative f″(x) is equal to the acceleration of the moving point at time x.
The third and fourth derivatives are defined similarly.
Definition. n-th derivative (or derivative n th order) functions y=f(x) called the derivative of it n-1-th derivative:
y (n) =(y (n-1))′, f (n) (x)=(f (n-1) (x))′.
Designations: y″′, y IV, y V etc.
Find an expression for the derivative of the exponential function \(y = (e^x)\), using the definition of the derivative.
Solution.
The initial steps are standard: first, write the increment of the function \(\Delta y\) corresponding to the increment of the argument \(\Delta x\): \[ (\Delta y = y\left((x + \Delta x) \right) - y\left(x \right) ) = ((e^(x + \Delta x)) - (e^x) ) = ((e^x)(e^(\Delta x)) - (e^x ) ) = ((e^x)\left(((e^(\Delta x)) - 1) \right).) \] The derivative is calculated as the limit of the increment ratio: \[ (y"\left(x \right ) = \lim\limits_(\Delta x \to 0) \frac((\Delta y))((\Delta x)) ) = (\lim\limits_(\Delta x \to 0) \frac((( e^x)\left(((e^(\Delta x)) - 1) \right)))((\Delta x)).) \] The function \(y = (e^x)\) in the numerator does not depend on Δ x and it can be taken out of the limit sign. Then the derivative takes the following form: \[ (y"\left(x \right) = (\left(((e^x)) \right)^\prime ) ) = ((e^x)\lim\limits_( \Delta x \to 0) \frac(((e^(\Delta x)) - 1))((\Delta x)).) \] Denote the resulting limit by \(L\) and calculate it separately. incidentally, \((e^0) = 1\) and, therefore, we can write \[ (L = \lim\limits_(\Delta x \to 0) \frac(((e^(\Delta x)) - 1))((\Delta x)) ) = (\lim\limits_(\Delta x \to 0) \frac(((e^(\Delta x)) - (e^0)))((\ Delta x)) = e"\left(0 \right),) \] that is, this limit is the value of the derivative of the exponential function at zero. Consequently, \ We have obtained a relation in which the desired derivative is expressed in terms of the function \(y = (e^x)\) itself and its derivative at the point \(x = 0\). Let us prove that \ To do this, recall that the number \(e\) is defined as an infinite limit as \ and the number \(e\) to the power \(\Delta x\) will be, respectively, equal to \[(e^(\ Delta x)) = \lim\limits_(n \to \infty ) (\left((1 + \frac((\Delta x))(n)) \right)^n).\] Next we apply the famous formula Newton's binomial and expand the expression under the limit sign in binomial series: \[(\left((1 + \frac((\Delta x))(n)) \right)^n) = \sum\limits_(k = 0)^n (C_n^k((\left( (\frac((\Delta x))(n)) \right))^k)) .\] ). In European and American textbooks, the number of combinations is denoted as \ Let's return to our limit \(L\), which can now be written as follows: \[ (L = \lim\limits_(\Delta x \to 0) \frac((( e^(\Delta x)) - 1))((\Delta x)) ) = (\lim\limits_(\Delta x \to 0) \frac((\lim\limits_(n \to \infty ) \ left[ (\sum\limits_(k = 0)^n (C_n^k((\left((\frac((\Delta x))(n)) \right))^k)) ) \right] - 1))((\Delta x)).) \] It is convenient for us to single out the first two terms in the binomial series: for \(k = 0\) and \(k = 1\). The result is \[ (L = \lim\limits_(\Delta x \to 0) \frac((\lim\limits_(n \to \infty ) \left[ (\sum\limits_(k = 0)^n (C_n^k((\left((\frac((\Delta x))(n)) \right))^k)) ) \right] - 1))((\Delta x)) ) = (\ lim\limits_(\Delta x \to 0) \frac((\lim\limits_(n \to \infty ) \left[ (C_n^0((\left((\frac((\Delta x))(n )) \right))^0) + C_n^1((\left((\frac((\Delta x))(n)) \right))^1) + \sum\limits_(k = 2)^ n (C_n^k((\left((\frac((\Delta x))(n)) \right))^k)) ) \right] - 1))((\Delta x)) ) = ( \lim\limits_(\Delta x \to 0) \frac((\lim\limits_(n \to \infty ) \left[ (1 + n \cdot \frac((\Delta x))(n) + \ sum\limits_(k = 2)^n (C_n^k((\left((\frac((\Delta x))(n)) \right))^k)) ) \right] - 1))( (\Delta x)) ) = (\lim\limits_(\Delta x \to 0) \frac((\Delta x + \lim\limits_(n \to \infty ) \sum\limits_(k = 2)^ n (C_n^k((\left((\frac((\Delta x))(n)) \right))^k)) ))((\Delta x)) ) = (\lim\limits_(\ Delta x \to 0) \left[ (1 + \frac(1)((\Delta x))\lim\limits_(n \to \infty ) \sum\limits_(k = 2)^n (C_n^k ((\left((\frac((\Delta x))(n)) \right))^k)) ) \right] ) = (1 + \lim\limits_(n \to \infty ) \left[ (\lim\limits_(\Delta x \to 0) \left((\sum\limits_(k = 2)^n (C_n^k\frac((((\left((\Delta x) \right)) ^(k - 1))))(((n^k)))) ) \right)) \right].) \] Obviously, the sum of the series tends to zero as \(\Delta x \to 0\) . Therefore, \(L = 1\). This means that the derivative of the exponential function \(y = (e^x)\) is equal to the function itself: \
When solving various problems of geometry, mechanics, physics and other branches of knowledge, it became necessary to use the same analytical process from a given function y=f(x) get a new function called derivative function(or simply derivative) of this function f(x) and are symbolized
The process by which a given function f(x) get a new function f"(x), called differentiation and it consists of the following three steps: 1) we give the argument x increment
x and determine the corresponding increment of the function
y = f(x+
x)-f(x); 2) make up the relation 
3) counting x permanent, and
x0, we find 
, which is denoted by f"(x), as if emphasizing that the resulting function depends only on the value x, at which we pass to the limit. Definition:
Derivative y "=f" (x)
given function y=f(x)
given x is called the limit of the ratio of the increment of the function to the increment of the argument, provided that the increment of the argument tends to zero, if, of course, this limit exists, i.e. finite. Thus, 
, or 
Note that if for some value x, for example when x=a, relation 
at
x0 does not tend to a finite limit, then in this case we say that the function f(x) at x=a(or at the point x=a) has no derivative or is not differentiable at a point x=a.
2. The geometric meaning of the derivative.
Consider the graph of the function y \u003d f (x), differentiable in the vicinity of the point x 0
f(x)
Let's consider an arbitrary straight line passing through the point of the graph of the function - the point A (x 0, f (x 0)) and intersecting the graph at some point B (x; f (x)). Such a straight line (AB) is called a secant. From ∆ABC: AC = ∆x; BC \u003d ∆y; tgβ=∆y/∆x .
Since AC || Ox, then ALO = BAC = β (as corresponding in parallel). But ALO is the angle of inclination of the secant AB to the positive direction of the Ox axis. Hence, tgβ = k is the slope of the straight line AB.
Now we will decrease ∆x, i.e. ∆x→ 0. In this case, point B will approach point A according to the graph, and the secant AB will rotate. The limiting position of the secant AB at ∆x → 0 will be the straight line (a), called the tangent to the graph of the function y \u003d f (x) at point A.
If we pass to the limit as ∆х → 0 in the equality tgβ =∆y/∆x, then we get 
or tg \u003d f "(x 0), since 
-angle of inclination of the tangent to the positive direction of the Ox axis 
, by definition of a derivative. But tg \u003d k is the slope of the tangent, which means that k \u003d tg \u003d f "(x 0).
So, the geometric meaning of the derivative is as follows:
Derivative of a function at a point x 0 equal to the slope of the tangent to the graph of the function drawn at the point with the abscissa x 0 .
3. Physical meaning of the derivative.
Consider the movement of a point along a straight line. Let the coordinate of the point at any time x(t) be given. It is known (from the course of physics) that the average speed over a period of time is equal to the ratio of the distance traveled during this period of time to the time, i.e.
Vav = ∆x/∆t. Let us pass to the limit in the last equality as ∆t → 0.
lim Vav (t) = (t 0) - instantaneous speed at time t 0, ∆t → 0.
and lim = ∆x/∆t = x "(t 0) (by the definition of a derivative).
So, (t) = x"(t).
The physical meaning of the derivative is as follows: the derivative of the functiony = f(x) at the pointx 0 is the rate of change of the functionf(x) at the pointx 0
The derivative is used in physics to find the speed from a known function of coordinates from time, acceleration from a known function of speed from time.
(t) \u003d x "(t) - speed,
a(f) = "(t) - acceleration, or
If the law of motion of a material point along a circle is known, then it is possible to find the angular velocity and angular acceleration during rotational motion:
φ = φ(t) - change in angle with time,
ω \u003d φ "(t) - angular velocity,
ε = φ"(t) - angular acceleration, or ε = φ"(t).
If the distribution law for the mass of an inhomogeneous rod is known, then the linear density of the inhomogeneous rod can be found:
m \u003d m (x) - mass,
x , l - rod length,
p \u003d m "(x) - linear density.
With the help of the derivative, problems from the theory of elasticity and harmonic vibrations are solved. Yes, according to Hooke's law
F = -kx, x – variable coordinate, k – coefficient of elasticity of the spring. Putting ω 2 \u003d k / m, we obtain the differential equation of the spring pendulum x "(t) + ω 2 x (t) \u003d 0,
where ω = √k/√m is the oscillation frequency (l/c), k is the spring rate (H/m).
An equation of the form y "+ ω 2 y \u003d 0 is called the equation of harmonic oscillations (mechanical, electrical, electromagnetic). The solution to such equations is the function
y = Asin(ωt + φ 0) or y = Acos(ωt + φ 0), where
A - oscillation amplitude, ω - cyclic frequency,
φ 0 - initial phase.
It is absolutely impossible to solve physical problems or examples in mathematics without knowledge about the derivative and methods for calculating it. The derivative is one of the most important concepts of mathematical analysis. We decided to devote today's article to this fundamental topic. What is a derivative, what is its physical and geometric meaning, how to calculate the derivative of a function? All these questions can be combined into one: how to understand the derivative?
Geometric and physical meaning of the derivative
Let there be a function f(x) , given in some interval (a,b) . The points x and x0 belong to this interval. When x changes, the function itself changes. Argument change - difference of its values x-x0 . This difference is written as delta x and is called argument increment. The change or increment of a function is the difference between the values of the function at two points. Derivative definition:
The derivative of a function at a point is the limit of the ratio of the increment of the function at a given point to the increment of the argument when the latter tends to zero.
Otherwise it can be written like this:
What is the point in finding such a limit? But which one:
the derivative of a function at a point is equal to the tangent of the angle between the OX axis and the tangent to the graph of the function at a given point.
The physical meaning of the derivative: the time derivative of the path is equal to the speed of the rectilinear motion.
Indeed, since school days, everyone knows that speed is a private path. x=f(t) and time t . Average speed over a certain period of time:
To find out the speed of movement at a time t0 you need to calculate the limit:
Rule one: take out the constant
The constant can be taken out of the sign of the derivative. Moreover, it must be done. When solving examples in mathematics, take as a rule - if you can simplify the expression, be sure to simplify .
Example. Let's calculate the derivative:
Rule two: derivative of the sum of functions
The derivative of the sum of two functions is equal to the sum of the derivatives of these functions. The same is true for the derivative of the difference of functions.
We will not give a proof of this theorem, but rather consider a practical example.
Find the derivative of a function:
Rule three: the derivative of the product of functions
The derivative of the product of two differentiable functions is calculated by the formula:
Example: find the derivative of a function:
Solution: 
Here it is important to say about the calculation of derivatives of complex functions. The derivative of a complex function is equal to the product of the derivative of this function with respect to the intermediate argument by the derivative of the intermediate argument with respect to the independent variable.
In the above example, we encounter the expression:
In this case, the intermediate argument is 8x to the fifth power. In order to calculate the derivative of such an expression, we first consider the derivative of the external function with respect to the intermediate argument, and then multiply by the derivative of the intermediate argument itself with respect to the independent variable.
Rule Four: The derivative of the quotient of two functions
Formula for determining the derivative of a quotient of two functions:
We tried to talk about derivatives for dummies from scratch. This topic is not as simple as it seems, so be warned: there are often pitfalls in the examples, so be careful when calculating derivatives.
With any question on this and other topics, you can contact the student service. In a short time, we will help you solve the most difficult control and deal with tasks, even if you have never dealt with the calculation of derivatives before.
It's very easy to remember.
Well, we will not go far, we will immediately consider the inverse function. What is the inverse of the exponential function? Logarithm:
In our case, the base is a number:
Such a logarithm (that is, a logarithm with a base) is called a “natural” one, and we use a special notation for it: we write instead.
What is equal to? Of course, .
The derivative of the natural logarithm is also very simple:
Examples:
- Find the derivative of the function.
- What is the derivative of the function?
Answers: The exponent and the natural logarithm are functions that are uniquely simple in terms of the derivative. Exponential and logarithmic functions with any other base will have a different derivative, which we will analyze later, after we go through the rules of differentiation.
Differentiation rules
What rules? Another new term, again?!...
Differentiation is the process of finding the derivative.
Only and everything. What is another word for this process? Not proizvodnovanie... The differential of mathematics is called the very increment of the function at. This term comes from the Latin differentia - difference. Here.
When deriving all these rules, we will use two functions, for example, and. We will also need formulas for their increments:
There are 5 rules in total.
The constant is taken out of the sign of the derivative.
If - some constant number (constant), then.
Obviously, this rule also works for the difference: .
Let's prove it. Let, or easier.
Examples.
Find derivatives of functions:
- at the point;
- at the point;
- at the point;
- at the point.
Solutions:
- (the derivative is the same at all points, since it's a linear function, remember?);
Derivative of a product
Everything is similar here: we introduce a new function and find its increment:
Derivative:
Examples:
- Find derivatives of functions and;
- Find the derivative of a function at a point.
Solutions:
Derivative of exponential function
Now your knowledge is enough to learn how to find the derivative of any exponential function, and not just the exponent (have you forgotten what it is yet?).
So where is some number.
We already know the derivative of the function, so let's try to bring our function to a new base:
To do this, we use a simple rule: . Then:
Well, it worked. Now try to find the derivative, and don't forget that this function is complex.
Happened?
Here, check yourself:
The formula turned out to be very similar to the derivative of the exponent: as it was, it remains, only a factor appeared, which is just a number, but not a variable.
Examples:
Find derivatives of functions:
Answers:
This is just a number that cannot be calculated without a calculator, that is, it cannot be written in a simpler form. Therefore, in the answer it is left in this form.
Note that here is the quotient of two functions, so we apply the appropriate differentiation rule:
In this example, the product of two functions:
Derivative of a logarithmic function
Here it is similar: you already know the derivative of the natural logarithm:
Therefore, to find an arbitrary from the logarithm with a different base, for example, :
We need to bring this logarithm to the base. How do you change the base of a logarithm? I hope you remember this formula:
Only now instead of we will write:
The denominator turned out to be just a constant (a constant number, without a variable). The derivative is very simple:
Derivatives of the exponential and logarithmic functions are almost never found in the exam, but it will not be superfluous to know them.
Derivative of a complex function.
What is a "complex function"? No, this is not a logarithm, and not an arc tangent. These functions can be difficult to understand (although if the logarithm seems difficult to you, read the topic "Logarithms" and everything will work out), but in terms of mathematics, the word "complex" does not mean "difficult".
Imagine a small conveyor: two people are sitting and doing some actions with some objects. For example, the first wraps a chocolate bar in a wrapper, and the second ties it with a ribbon. It turns out such a composite object: a chocolate bar wrapped and tied with a ribbon. To eat a chocolate bar, you need to do the opposite steps in reverse order.
Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then we will square the resulting number. So, they give us a number (chocolate), I find its cosine (wrapper), and then you square what I got (tie it with a ribbon). What happened? Function. This is an example of a complex function: when, in order to find its value, we do the first action directly with the variable, and then another second action with what happened as a result of the first.
In other words, A complex function is a function whose argument is another function: .
For our example, .
We may well do the same actions in reverse order: first you square, and then I look for the cosine of the resulting number:. It is easy to guess that the result will almost always be different. An important feature of complex functions: when the order of actions changes, the function changes.
Second example: (same). .
The last action we do will be called "external" function, and the action performed first - respectively "internal" function(these are informal names, I use them only to explain the material in simple language).
Try to determine for yourself which function is external and which is internal:
Answers: The separation of inner and outer functions is very similar to changing variables: for example, in the function
- What action will we take first? First we calculate the sine, and only then we raise it to a cube. So it's an internal function, not an external one.
And the original function is their composition: . - Internal: ; external: .
Examination: . - Internal: ; external: .
Examination: . - Internal: ; external: .
Examination: . - Internal: ; external: .
Examination: .
we change variables and get a function.
Well, now we will extract our chocolate - look for the derivative. The procedure is always reversed: first we look for the derivative of the outer function, then we multiply the result by the derivative of the inner function. For the original example, it looks like this:
Another example:
So, let's finally formulate the official rule:
Algorithm for finding the derivative of a complex function:
It seems to be simple, right?
Let's check with examples:
Solutions:
1) Internal: ;
External: ;
2) Internal: ;
(just don’t try to reduce by now! Nothing is taken out from under the cosine, remember?)
3) Internal: ;
External: ;
It is immediately clear that there is a three-level complex function here: after all, this is already a complex function in itself, and we still extract the root from it, that is, we perform the third action (put chocolate in a wrapper and with a ribbon in a briefcase). But there is no reason to be afraid: anyway, we will “unpack” this function in the same order as usual: from the end.
That is, first we differentiate the root, then the cosine, and only then the expression in brackets. And then we multiply it all.
In such cases, it is convenient to number the actions. That is, let's imagine what we know. In what order will we perform actions to calculate the value of this expression? Let's look at an example:
The later the action is performed, the more "external" the corresponding function will be. The sequence of actions - as before:
Here the nesting is generally 4-level. Let's determine the course of action.
1. Radical expression. .
2. Root. .
3. Sinus. .
4. Square. .
5. Putting it all together:
DERIVATIVE. BRIEFLY ABOUT THE MAIN
Function derivative- the ratio of the increment of the function to the increment of the argument with an infinitesimal increment of the argument:
Basic derivatives:
Differentiation rules:
The constant is taken out of the sign of the derivative:
Derivative of sum:
Derivative product:
Derivative of the quotient:
Derivative of a complex function:
Algorithm for finding the derivative of a complex function:
- We define the "internal" function, find its derivative.
- We define the "external" function, find its derivative.
- We multiply the results of the first and second points.
